Conceptual doubt in angular velocity

I am having a few doubts while writing relationship between angular speed and linear speed ,in the reference frame of a moving observer .

Suppose at an instant, car A is heading towards car B with velocity 'v'. Car B is moving towards right with velocity 'u' ,i.e along positive x-axis . The instantaneous distance between A and B is 's' .

What is the relationship between the different parameters at this instant ?

2. Relevant equations

3. The attempt at a solution

In the reference frame of car A ,the car B is rotating with angular velocity ω ,and translating with velocity ucosθ-v .

So , uT = sω where uT=usinθ is the tangential speed of car B .

Now,my confusion is angular speed is the time rate of change of angle .But which angle ? Is it θ or α or something else?

If it is θ then $$ s\dotθ = -usinθ $$ . If it is α ,then $$ s\dotα = usinθ $$ .

While dealing with angular speeds,I have difficulty in determining the angle ,whose time rate of change is ω.

I am not sure if what I have written makes sense .

I would be grateful if somebody could reflect his/her thoughts on this .

"Which angle" is really up to you, in two ways. First, the location of the "zero" angle is completely arbitrary, it can be the x-axis as depicted, the y-axis, or any other position. Second, you need to choose the "positive" direction. The choice is usually made so that the direction of the angular velocity vector will complete a right-handed triad with the x and y-axes. See http://en.wikipedia.org/wiki/Right-hand_rule . In 2D motion, this simply means that the positive direction is counter-clockwise.

Since A is heading towards B, A's velocity has no affect on the bearing of B from A. It's not clear how you are defining alpha and theta. If they are defined by the relative positions of A and B, as in the diagram, at any time t, then both your equations are valid. ##\dot \alpha = -\dot \theta##.

"Which angle" is really up to you, in two ways. First, the location of the "zero" angle is completely arbitrary, it can be the x-axis as depicted, the y-axis, or any other position. Second, you need to choose the "positive" direction. The choice is usually made so that the direction of the angular velocity vector will complete a right-handed triad with the x and y-axes. See http://en.wikipedia.org/wiki/Right-hand_rule . In 2D motion, this simply means that the positive direction is counter-clockwise.

Hi Voko...

Does that mean I can arbitrarily choose any line in the X-Y plane and consider angle between that line and AB ( say,β ) .Now $$ u_T =s\dot \beta $$ or $$ u_T = - s\dotβ $$ ,depending on whether I choose positive to be clockwise or anticlockwise .

It's not clear how you are defining alpha and theta. If they are defined by the relative positions of A and B, as in the diagram, at any time t, then both your equations are valid. ##\dot \alpha = -\dot \theta##.

Alpha is the angle which line joining A to B makes with Y-axis .Theta is the angle with the X-axis .

I believe its the "bearing" term that is the cause of problem. I quote Wikipedia :

In navigational terms, "bearing" is perhaps more usually the angle between our forward direction, and the direction from us to another object. It typically refers to the direction of, some object, as seen by us, compared to our current heading. In other words, it's simply the angle between our forward direction, and a line towards the object in question.

what haruspex meant was that direction of velocity of A and displacement vector from A to B has angle zero between them. I hope this helps!!!

Does this mean - If the velocity of A is towards +y axis (v) instead of towards B ,the relation will be

$$s\dot\theta = -(usin\theta + vcos\theta)$$ Is it correct ?

I meant that, instantaneously, the fact that A is moving has no affect on alpha, because the direction of A's movement happens to be towards B.

Alpha is the angle which line joining A to B makes with Y-axis .Theta is the angle with the X-axis .

Yes, but that could be just the instantaneous condition. Since we are concerned with changes in alpha it's important to be clear that that definition applies at all times, even as A and B move. For example, it could have been that A is on a constant heading and only happens to be heading towards B at that instant. In that case alpha could be defined as that constant heading, and by definition its rate of change would have been zero.

Staff: Mentor

In my judgement, the easiest way to do this problem is to use vectors. Here is an example:

velocity vector for car A[itex]=vcosθ\vec{i}+vsinθ\vec{j}[/itex]
velocity vector for car B [itex]=u\vec{i}[/itex]
position vector of car A [itex]=-s\sinθ\vec{j}[/itex]
position vector of car B [itex]=s\cosθ\vec{i}[/itex]

Relative velocity vector of car B relative to car A [itex]=\vec{V_r}=u\vec{i}-(vcosθ\vec{i}+vsinθ\vec{j})=(u-vcosθ)\vec{i}-vsinθ\vec{j}[/itex]
Relative position vector of car B relative to car A [itex]=s\cosθ\vec{i}+s\sinθ\vec{j}=s(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
Unit vector in direction of car B relative to car A [itex]=\vec{U_r}=(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
Component of relative velocity vector of car B relative to car A in the direction of car B relative to car A = [itex](\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(ucosθ-v)\vec{U_r}[/itex]
Component of relative velocity vector of car B relative to car A in the direction perpendicular to the direction of car B relative to car A = [itex]\vec{V_r}-(\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(u-vcosθ)\vec{i}-vsinθ\vec{j}-(ucosθ-v)(\cosθ\vec{i}+\sinθ\vec{j})=u\sin^2θ\vec{i}-u\sinθcosθ\vec{j}=usinθ(sinθ\vec{i}-cosθ\vec{j})[/itex]
The magnitude of this perpendicular relative velocity component is just u sinθ.

Therefore, the angular velocity of car B relative to car A is given by:

Staff: Mentor

##\vec r## is the position vector, ##\vec v## is the velocity. θ is the angle of ##\vec v## with respect to the direction of ##\vec r##. See picture.

If you determine the position vector and the relative velocity of B with respect to A you get the angular velocity of B with respect to A or vice versa from the cross product.

ehild

Yes. Use of the cross product is another good way to get the same result. Also, once I saw what the answer was and considered it in comparison with the figure, I could have kicked myself. I realized that the answer could have been written down almost on inspection. All that really needed to be done was to resolve u into components parallel and perpendicular to the line joining A and B; v was already aligned along the parallel direction.

In my judgement, the easiest way to do this problem is to use vectors. Here is an example:

velocity vector for car A[itex]=vcosθ\vec{i}+vsinθ\vec{j}[/itex]
velocity vector for car B [itex]=u\vec{i}[/itex]
position vector of car A [itex]=-s\sinθ\vec{j}[/itex]
position vector of car B [itex]=s\cosθ\vec{i}[/itex]

Relative velocity vector of car B relative to car A [itex]=\vec{V_r}=u\vec{i}-(vcosθ\vec{i}+vsinθ\vec{j})=(u-vcosθ)\vec{i}-vsinθ\vec{j}[/itex]
Relative position vector of car B relative to car A [itex]=s\cosθ\vec{i}+s\sinθ\vec{j}=s(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
Unit vector in direction of car B relative to car A [itex]=\vec{U_r}=(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
Component of relative velocity vector of car B relative to car A in the direction of car B relative to car A = [itex](\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(ucosθ-v)\vec{U_r}[/itex]
Component of relative velocity vector of car B relative to car A in the direction perpendicular to the direction of car B relative to car A = [itex]\vec{V_r}-(\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(u-vcosθ)\vec{i}-vsinθ\vec{j}-(ucosθ-v)(\cosθ\vec{i}+\sinθ\vec{j})=u\sin^2θ\vec{i}-u\sinθcosθ\vec{j}=usinθ(sinθ\vec{i}-cosθ\vec{j})[/itex]
The magnitude of this perpendicular relative velocity component is just u sinθ.

Therefore, the angular velocity of car B relative to car A is given by:

[tex]ω=\frac{u}{s}sinθ[/tex]

Thanks Chet for your excellent input .This is a very good and systematic way of handling the problem .You have given a very nice insight in using vectors to deal with the problems .

"Car A moves uniformly with velocity ‘v’ so that its velocity is continually towards car B which in turn moves uniformly with velocity ‘u’ in x-direction (u<v). At t=0 , car A is at origin and car B points towards origin at a distance L . After how much time car B would catch up with car A ?"

This is one way of approaching the problem .

At any instant , let the distance between the cars be ‘s ‘ and the angle between the two velocities be θ .

Then ds/dt = -(v-ucos θ) .

Integrating L = vT – ∫ucosθdt (1)

uT = ∫vcosθdt (2)

So , T = (vl)/(v2 – u2)

Is there any other way of thinking about this problem ? Can this problem be done vectorially or some other approach ? A longer method is perfectly alright .

Staff: Mentor

"Car A moves uniformly with velocity ‘v’ so that its velocity is continually towards car B which in turn moves uniformly with velocity ‘u’ in x-direction (u<v). At t=0 , car A is at origin and car B points towards origin at a distance L . After how much time car B would catch up with car A ?"

This is one way of approaching the problem .

At any instant , let the distance between the cars be ‘s ‘ and the angle between the two velocities be θ .

Then ds/dt = -(v-ucos θ) .

Integrating L = vT – ∫ucosθdt (1)

uT = ∫vcosθdt (2)

So , T = (vl)/(v2 – u2)

Is there any other way of thinking about this problem ? Can this problem be done vectorially or some other approach ? A longer method is perfectly alright .

You can't do it that way because θ is changing with time.
I recommend using polar coordinates centered at car A, and starting with the results from the previous example.